Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        StringTokenizer st = new StringTokenizer(br.readLine());
        double[][] dp = new double[n + 1][n + 2];

        for (int i = 1; i <= n; i++) {
            double p = Double.parseDouble(st.nextToken());

            if (i == 1) {
                dp[1][0] = 1 - p;
                dp[1][1] = p;
            }
            else {
                dp[i][0] = dp[i - 1][0] * (1 - p);

                for (int j = 1; j < i; j++) {
                    dp[i][j] = dp[i - 1][j] * (1 - p) + dp[i - 1][j - 1] * p;
                    dp[i][j + 1] = dp[i - 1][j] * p + dp[i - 1][j + 1] * (1 - p);
                }
            }
        }

        double sum = 0;

        for (int i = n / 2 + 1; i <= n; i++)
            sum += dp[n][i];

        System.out.println(sum);
    }
}